Simple Pendulum — Time period, effective length

Q36.A particle oscillates along the x-axis according to the law, x(t) = x0 sin2 ( 2t ) where x0 = 1 m . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph (1) (2) (3) (4)

Q28.Two bodies A and B of equal mass are suspended from two massless springs of spring constant k1 and k2 , respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is (1) k1 (2) k2 √k1k2 (3) (4) k2 k1 √k2k1

Q41.A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is yπ × 10−2 s, where the value of y is (Acceleration due to gravity, g = 10 m/s2 , density of water = 103 kg/m3 ) 2025 (23 Jan Shift 1) JEE Main Previous Year Paper (1) 6 (2) 2 (3) 4 (4) 1

Q36.A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm . If D and d are the total distance and displacement covered by the particle in 12.5 s , then D is d (1) 16 (2) 10 5 (3) 15 (4) 25 4

Q35.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Knowing initial position x0 and initial momentum p0 is enough to determine the position and momentum at any time t for a simple harmonic motion with a given angular frequency ω. Reason (R): The amplitude and phase can be expressed in terms of x0 and p0 . In the light of the above statements, choose the correct answer from the options given below : (1) (A) is false but (R) is true (2) (A) is true but (R) is false (3) Both (A) and (R) are true but (R) is NOT the (4) Both (A) and (R) are true and (R) is the correct correct explanation of (A) explanation of (A)

Q24.The displacement of a particle executing SHM is given by x = 10 sin (wt + π3 )m. The time period of motion is 3.14 s. The velocity of the particle at t = 0 is ______ m/s.